Physics, asked by venkatsagi2001, 1 year ago

Q 44 Consider a planet whose mass is three times that of Earth's and radius is 3/2 times of Earth's. Find the difference between acceleration due to gravity of
planet to the acceleration due to gravity on Earth.
Note:
- Mass of earth = 6 * 10
24
kg
- Diameter of earth = 12800 km
-11
- G = 6.67 * 10 Nm/s?
- Accceleration due to gravity = 10 m/s?​

Answers

Answered by vaibhavsemwal
0

Answer:

Acceleration due to gravity at other planet is 13.33m/s^2.

Explanation:

We know that acceleration due to gravity is given by, g_e= \frac{GM_e}{R_e^2}=10m/s^2

where, G is the gravitational constant = 6.67*10^{-11}

M_e is the mass of the earth = 6*10^{24} Kg

R_e is the radius of the earth = \frac{128*10^{5}}{2}=64*10^{5}m

Acceleration due to gravity at other planet,  g_p = \frac{G(3.M_E)}{(\frac{3}{2}R_e )^2}

\implies g_p=\frac{4}{3} \frac{GM_e}{R_e^2}

\implies g_p=\frac{4}{3} *10

\implies g_p=13.33m/s^2

#SPJ2

   

Answered by syed2020ashaels
0

Answer:

Acceleration due to gravity of the planet is 13.33m/sec^2

Explanation:

  • Gravity is the force with which the earth attracts a body towards its centre.
  • Now as we know that
    g = \frac{GM}{r^2}\\g=10 m/sec^2
  • Now according to the question a planet which is \frac{3}{2} times the radius of earth and the mass is 3 times that of earth
  • So the of that planet is
    g_{p} = \frac{3GM}{\frac{9r^2}{4}}\\g_{p} = \frac{12}{9}g\\
  • So the difference is
    G = g*\frac{12}{9}-1\\G = g*\frac{3}{9}\\G = 10*1.3333\\G=13.3333 m/sec^2
    #SPJ2
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